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In triangle $ABC,$ $AB = 15,$ $BC = 10,$ and $AC = 12.$ Find the length of the shortest altitude in this triangle.

 Apr 8, 2024
 #1
avatar+14943 
+1

Find the length of the shortest altitude in this triangle.

 

\(a_a:a_b:a_c=\frac{1}{a}:\frac{1}{b}:\frac{1}{c}\\ a_a:a_b:a_c=\frac{1}{10}:\frac{1}{12}:\frac{1}{15}\\\)

\({\color{blue}The\ shortest\ altitude}\ in\ this\ triangle\ \color{blue}is\ a_c.\\ \overline{AC}=12\ ,\ \overline{BC}=10\ ,\ \overline{AB}=15\)

 

            C

             |

             \(a_c\)                      Triangle ABC

             |

A     x    .     15-x    B

 

\( a_c\ ^2=12^2-x^2=10^2-(15-x)^2\\ 144-x^2=100-(225-30x+x^2)\\ 144-100+225=30x\\ x=8.9\overline 6\)

\(a_c\ ^2=12^2-8.9\overline 6\ ^2\\ a_c=\sqrt{12^2-8.9\overline 6\ ^2}\\ \color{blue}a_c=7.97489\)

 

laugh !

 Apr 8, 2024
edited by asinus  Apr 10, 2024
 #2
avatar+128794 
+1

Semi-perimeter =  [ 15 + 10 + 12 ]  / 2 =  18.5

 

Area  = sqrt  [18.5 * 3.5 * 6.5 * 8.5 ]  ≈  59.8

 

Area =  (1/2) (longest side )( altitude to this side)

 

 59.8  = (1/2) (15)  (altitude to this side)

 

59.8 / 7.5  ≈  7.97  =  shortest altitude

 

cool cool cool

 Apr 9, 2024

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